General Solution Where All Airplanes Survive
If we don't want to destroy any planes, this problem leads to a nifty insight - not surprising since it was shared by Martin Gardner. I will answer his variant asking for the number of planes needed so the treatment is more general.
The general template for the strategy is similar to Chris Cudmore's answer ( https://puzzling.stackexchange.com/a/117441/81255 ) : there is some send-off help where the leading airplane receives fuel, followed by a solo leg for the leader, finished by a last-stretch help where the leading airplane receives some fuel again.
A Note on Fuel-Transfer
Say there are $n$ planes flying: a chosen leader, $L$, to make the full journey around the world, and $(n-1)$ helpers. Since fuel-transfer takes $0$ time between any 2 airplanes, one can make infinitesimal transfers between any two airplanes in quick succession. Quick succession with $0$ time in between is effectively simultaneous fuel-transfer. So we can assume that many airplanes can send/receive fuel simultaneously and continuously from any number of airplanes.
Units of Measurement
Let's measure distance in fraction of great-circle length, and fuel in fraction of tank-capacity. Since time and velocity aren't specified in the problem, we can simply measure time as distance. So we can say "a plane loses fuel at a rate of $2x$" to mean "if the plane travels a distance $x$ (fraction of great-circle), its fuel reduces by $2x$ (fraction of tank-capacity)"
Burn Factor
When an airplane is helping, it loses fuel at a rate of $2bx$ for some $b>1$, since it is spending fuel for itself and at least one other airplane. Let's call $b$ the burn factor. For example, if a single helper is helping $(n-1)$ other airplanes, then $n$ airplanes are flying on effectively 1 tank and the burn factor, $b$, is $n$. The burn factor here is essentially the number of planes the helper's tank is fueling.
Strategy
On the send-off stretch, the first helper plane fuels all other planes for a distance $r_1$ and then returns to refuel, while the the second helper plane takes over and fuels the remaining for a distance $r_2$ and returns, and so on, with airplane $i$ fueling the remaining pack for a distance $r_i$ before returning to refuel, till only the leader, $L$, is left.
$L$ flies solo for a distance.
The refueled helpers fly in the opposite direction, rewinding the send-off pattern: the first refueled helper rewinds the tape of the last returned helper. Clearly they can help $L$ the same distance on the last-stretch as in the send-off stretch and all planes land.
Analysis
We just need to consider the total send-off helping distance since the last-stretch helping distance is the same.
The first helper will lose fuel at the rate of $2b_1 x$ when helping, with a burn factor $b_1 = n$. Helper plane $i$ will lose fuel at the rate of $2 b_i x$ when it is helping, with $b_i = b_{i-1} - 1$ since it is fueling one less plane than the previous helper. Every helper plane loses fuel at the rate of $2x$ when returning to refuel.
Helper plane $i$ will have a full tank when it starts helping, when helper $(i-1)$ turns back to refuel. It has one less plane to help compared to $(i-1)$, but it will have to travel a longer distance back to the island to refuel.
If every helper helps maximally, it will have just enough fuel to reach the island when it turns back. The first helper will have $1 - 2 b_1 r_1$ fuel left in the tank when it turns, which should be exactly $2r_1$ to make it back.The second helper will need to have $2(r_1 + r_2)$ fuel to make it back, and so on:
$$ \begin{array}{ll}
1 - 2 b_1 r_1 = 2 r_1, \\
1 - 2b_2 r_2 = 2 (r_1 + r_2) = (1 - 2b_1 r_1) + 2r_2, \\
\vdots \\
1 - 2b_{n-1} r_{n-1} = 2(r_1 + r_2 + \cdots + r_{n-1}) = (1 - 2b_{n-2}r_{n-2}) + 2r_{n-1},
\end{array}
$$
or, more succinctly,
$$
1 - 2b_i r_i = 1 - 2b_{i-1} r_{i-1} + 2 r_i, \quad 2 \leq i \leq n-1.
$$
Since $b_i = b_{i-1} - 1$ and $b_1 = n$, it is straightforward to see that
$$ r_i = \frac{1}{2(b_1 + 1)} = \frac{1}{2(n+1)} \forall i < n. $$
So every helper plane pushes itself and the remaining pack the same distance forward. Since there are $n-1$ helpers, the total send-off helping distance is
$$ \sum_{i=1} ^{n-1} r_i = \frac{n-1}{2(n+1)}. $$
The last-stretch helping distance is identical since it is just a tape-rewind of the send-off, resulting in a total helping distance of
$$ \frac{n-1}{n+1}. $$
Clearly this has to be at least $1/2$ for $L$ to be able to fly the rest by itself. So
$$ \frac{n-1}{n+1} \geq \frac{1}{2} \implies n \geq 3. $$
Addendum 1
Interestingly, when $n=3$, all planes will be empty upon reaching the finish-line: on the last stretch, $L$ will be met just in time by one helper exactly when it is empty, and the first helper will be met by the second exactly when it is empty. So the total fuel used in this case will be 5 tanks.
Addendum 2
A seemingly viable alternative, where all helpers simultaneously help $L$ (but not each other), is actually never viable for any $n$, since the ratio of how much a helper's fuel goes into flying itself versus flying another plane is worse (larger).
Hope this amuses/helps!
